题目
Problem
(a) By considering the binomial expansion of ( z + z − 1 ) 4 \big(z+z^{-1}\big)^4 ( z + z − 1 ) 4 z = cos θ + i sin θ z=\cos\theta+\mathrm{i}\sin\theta z = cos θ + i sin θ cos 4 θ = 1 8 ( cos 4 θ + 4 cos 2 θ + 3 ) \cos^4\theta=\frac{1}{8}(\cos4\theta+4\cos2\theta+3) cos 4 θ = 8 1 ( cos 4 θ + 4 cos 2 θ + 3 )
(5)
(b) Use the substitution x = sin θ x=\sin\theta x = sin θ ∫ 0 1 2 ( 1 − x 2 ) 3 2 d x \displaystyle\int_0^{\frac{1}{2}}\big(1-x^2\big)^{\frac{3}{2}}\,\mathrm{d}x ∫ 0 2 1 ( 1 − x 2 ) 2 3 d x
(3)
题目中文翻译
(a) 通过考虑 ( z + z − 1 ) 4 \big(z+z^{-1}\big)^4 ( z + z − 1 ) 4 z = cos θ + i sin θ z=\cos\theta+\mathrm{i}\sin\theta z = cos θ + i sin θ cos 4 θ = 1 8 ( cos 4 θ + 4 cos 2 θ + 3 ) \cos^4\theta=\frac{1}{8}(\cos4\theta+4\cos2\theta+3) cos 4 θ = 8 1 ( cos 4 θ + 4 cos 2 θ + 3 )
(5)
(b) 用代换 x = sin θ x=\sin\theta x = sin θ ∫ 0 1 2 ( 1 − x 2 ) 3 2 d x \displaystyle\int_0^{\frac{1}{2}}\big(1-x^2\big)^{\frac{3}{2}}\,\mathrm{d}x ∫ 0 2 1 ( 1 − x 2 ) 2 3 d x
(3)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
展开
把 z + z − 1 z+z^{-1} z + z − 1 2 cos θ 2\cos\theta 2 cos θ
答题过程
展开
Since ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 z − 1 = cos θ − i sin θ z^{-1}=\cos\theta-\mathrm{i}\sin\theta z − 1 = cos θ − i sin θ z + z − 1 = 2 cos θ z+z^{-1}=2\cos\theta z + z − 1 = 2 cos θ
( z + z − 1 ) 4 = z 4 + 4 z 2 + 6 + 4 z − 2 + z − 4 . (z+z^{-1})^4=z^4+4z^2+6+4z^{-2}+z^{-4}. ( z + z − 1 ) 4 = z 4 + 4 z 2 + 6 + 4 z − 2 + z − 4 . Pairing opposite powers gives
16 cos 4 θ = ( z 4 + z − 4 ) + 4 ( z 2 + z − 2 ) + 6. 16\cos^4\theta=(z^4+z^{-4})+4(z^2+z^{-2})+6. 16 cos 4 θ = ( z 4 + z − 4 ) + 4 ( z 2 + z − 2 ) + 6. By de Moivre’s theorem, z k + z − k = 2 cos ( k θ ) z^k+z^{-k}=2\cos(k\theta) z k + z − k = 2 cos ( k θ ) 16 16 16
cos 4 θ = 1 8 ( cos 4 θ + 4 cos 2 θ + 3 ) . \cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3). cos 4 θ = 8 1 ( cos 4 θ + 4 cos 2 θ + 3 ) .
(b)
解法一
思路
展开
令 x = sin θ x=\sin\theta x = sin θ d x \mathrm{d}x d x cos θ \cos\theta cos θ cos 4 θ \cos^4\theta cos 4 θ
答题过程
展开
Here d x = cos θ d θ \mathrm{d}x=\cos\theta\,\mathrm{d}\theta d x = cos θ d θ θ = 0 \theta=0 θ = 0 θ = π / 6 \theta=\pi/6 θ = π /6 cos θ > 0 \cos\theta>0 cos θ > 0 ( 1 − x 2 ) 3 / 2 = cos 3 θ (1-x^2)^{3/2}=\cos^3\theta ( 1 − x 2 ) 3/2 = cos 3 θ
∫ 0 1 / 2 ( 1 − x 2 ) 3 / 2 d x = ∫ 0 π / 6 cos 4 θ d θ = 1 8 [ 1 4 sin 4 θ + 2 sin 2 θ + 3 θ ] 0 π / 6 = 1 8 ( 3 8 + 3 + π 2 ) = 9 3 64 + π 16 . \begin{align*}
\int_0^{1/2}(1-x^2)^{3/2}\,\mathrm{d}x
=&\,\int_0^{\pi/6}\cos^4\theta\,\mathrm{d}\theta\\
=&\,\frac18\bigg[\frac14\sin4\theta+2\sin2\theta+3\theta\bigg]_0^{\pi/6}\\
=&\,\frac18\bigg(\frac{\sqrt3}{8}+\sqrt3+\frac\pi2\bigg)\\
=&\,\boxed{\frac{9\sqrt3}{64}+\frac\pi{16}}.
\end{align*} ∫ 0 1/2 ( 1 − x 2 ) 3/2 d x = = = = ∫ 0 π /6 cos 4 θ d θ 8 1 [ 4 1 sin 4 θ + 2 sin 2 θ + 3 θ ] 0 π /6 8 1 ( 8 3 + 3 + 2 π ) 64 9 3 + 16 π . 